Question

Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy. x5y2 − x4y + 2xy3 = 0

Answer 1

Answer:

dx/dy =  (x^4 -  2x^5y -  6xy^2) /  (5x^4y^2 - 4x^3y +  2y^3).

Step-by-step explanation:

x^5y^2 − x^4y + 2xy^3 = 0

Applying the Product and Chain Rules:

y^2*5x^4*dx/dy + 2y*x^5 - (y*4x^3*dx/dy + x^4) + (y^3* 2*dx/dy + 3y^2*2x) =0

Separating the terms with derivatives:

y^2*5x^4*dx/dy - y*4x^3*dx/dy + y^3* 2*dx/dy =  x^4 -  2y*x^5 -  3y^2*2x

dx/dy =  (x^4 -  2x^5y -  6xy^2) /  (5x^4y^2 - 4x^3y +  2y^3)

Answer 2

Answer:

$\frac{dx}{dy}=\frac{-2x^5y+x^4-6xy^2}{5x^4y^2-4x^3y+2y^3}$

Step-by-step explanation:

The given equation is

$x^5y^2-x^4y+2xy^3=0$

Differentiate with respect to y.

$\frac{d}{dy}(x^5y^2)-\frac{d}{dy}(x^4y)+\frac{d}{dy}(2xy^3)=0$

Using product rule we get

$x^5\frac{d}{dy}(y^2)+y^2\frac{d}{dy}(x^5)-x^4\frac{d}{dy}(y)-y\frac{d}{dy}(x^4)+2x\frac{d}{dy}(y^3)+2y^3\frac{d}{dy}(x)=0$           $(fg)'=fg'+gf'$

$x^5(2y)+y^2(5x^4\frac{dx}{dy})-x^4(1)-y(4x^3\frac{dx}{dy})+2x(3y^2)+2y^3\frac{dx}{dy}=0$

$2x^5y+5x^4y^2\frac{dx}{dy}-x^4-4x^3y\frac{dx}{dy}+6xy^2+2y^3\frac{dx}{dy}=0$

Isolate $\frac{dx}{dy}$ terms on left side.

$5x^4y^2\frac{dx}{dy}-4x^3y\frac{dx}{dy}+2y^3\frac{dx}{dy}=-2x^5y+x^4-6xy^2$

$(5x^4y^2-4x^3y+2y^3)\frac{dx}{dy}=-2x^5y+x^4-6xy^2$

Isolate $\frac{dx}{dy}$ term.

$\frac{dx}{dy}=\frac{-2x^5y+x^4-6xy^2}{5x^4y^2-4x^3y+2y^3}$

Therefore the value of $\frac{dx}{dy}$ is $\frac{-2x^5y+x^4-6xy^2}{5x^4y^2-4x^3y+2y^3}$.